A bob of mass ma suspended by a massle3ss inextensible string is rotated in a vertical circle of radius r as shoen in the figure. Obtain an expenssion for speeds and kinetic energies at A, B and C, Comment on the nature of trajectory of the bob after it reaches the point C.

Explanation

Let horizontal velocity at lowest point A be `V_(A)` such that string becomes stack only on reaching highest point C. Taking the potential energy of the system to be zero at A, the total mechanical energy, E
At A :
`E=(1)/(2)mv_(A)^(2)" ...(i) "(E=KE+PE=KE" as "PE=0)`
`T_(A)-mg=(mv_(A)^(2))/(r)" ...(ii) (Using Newton.s IInd Law)"`
Where `T_(A)` is the tension in the string at A.
At the highest point C, the string slackens, so `T_(c)=0`
At C :
`E=(1)/(2)mv_(C)^(2)+2mgr" ...(iii) "("where "v_(C)" is the velocity at C")`
`mg=(mv_(C)^(2))/(r )" ....(iv) (Newton.s IInd law)"`
From equation (iv), `(1)/(2)mv_(C)^(2)=(1)/(2)mgr`
Put this value in eq. (iii)
`E=(1)/(2)mgr+2mgr`
`E=(5)/(2)mgr" ...(v) "("This is also KE at A as "PE_(A)=0)`
From equations (i) & (v)
`(1)/(2)mv_(A)^(2)=(5)/(2)mgr`
`v_(A)=sqrt(5gr)" ...(vi)"`
`"Velocity at A"=sqrt(5gR)`
`"Kinetic energy at A"=(5)/(2)mgr" ....(vii)"`
From equation (iv), `v_(c)=sqrt(gr)" ....(viii)"`
`"Kinetic energy at C"=(1)/(2)mgr" ...(ix)"`
At B :
`E=(1)/(2)mv_(B)^(2)+mgr" ...(x) "("Where "v_(B)" is the velocity at B")`
This should be same as the energy at A.
So, `(1)/(2)mv_(B)^(2)+mgr=(1)/(2)mv_(A)^(2)=(5)/(2)mgr" (Using equation (vii))"`
`(1)/(2)mv_(B)^(2)=(3)/(2)mgr`
`rArr" "v_(B)=sqrt(3gr)" ...(xi)"`
`"Kinetic energy at B"=(3)/(2)mgr" ...(xii)"`
At C, the string becomes slack and velocity of bob is horizontal and to the left, Bob continues on its circular path and completes the revolution.